3.12.87 \(\int \frac {1+k^2 x^4}{\sqrt {(1-x^2) (1-k^2 x^2)} (-1+k^2 x^4)} \, dx\) [1187]

Optimal. Leaf size=87 \[ -\frac {\text {ArcTan}\left (\frac {(-1+k) x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 (-1+k)}+\frac {\text {ArcTan}\left (\frac {(1+k) x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{-1-k} \]

[Out]

-arctan((-1+k)*x/(1+(-k^2-1)*x^2+k^2*x^4)^(1/2))/(-2+2*k)+arctan((1+k)*x/(1+k*x^2+(1+(-k^2-1)*x^2+k^2*x^4)^(1/
2)))/(-1-k)

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Rubi [A]
time = 0.60, antiderivative size = 81, normalized size of antiderivative = 0.93, number of steps used = 14, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {1976, 6857, 1117, 2098, 1224, 1712, 210, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {(1-k) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{2 (1-k)}-\frac {\text {ArcTan}\left (\frac {(k+1) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{2 (k+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + k^2*x^4)/(Sqrt[(1 - x^2)*(1 - k^2*x^2)]*(-1 + k^2*x^4)),x]

[Out]

-1/2*ArcTan[((1 - k)*x)/Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]]/(1 - k) - ArcTan[((1 + k)*x)/Sqrt[1 - (1 + k^2)*x^2
+ k^2*x^4]]/(2*(1 + k))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1117

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(
4*c))], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1224

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[
a + b*x^2 + c*x^4], x], x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x] /; Fr
eeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1712

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[
A, Subst[Int[1/(d - (b*d - 2*a*e)*x^2), x], x, x/Sqrt[a + b*x^2 + c*x^4]], x] /; FreeQ[{a, b, c, d, e, A, B},
x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 1976

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[u*(a*c*e + (b*c
+ a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2098

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(PP /. x ->
x^2)^p*Q^q, x], x] /;  !SumQ[NonfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x^2] && PolyQ[Q, x] && ILtQ[p,
0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1+k^2 x^4}{\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (-1+k^2 x^4\right )} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \left (\frac {1}{\sqrt {1-x^2} \sqrt {1-k^2 x^2}}+\frac {2}{\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (-1+k^2 x^4\right )}\right ) \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \sqrt {1-k^2 x^2}} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (-1+k^2 x^4\right )} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \left (-\frac {1}{2 \sqrt {1-x^2} \left (1-k x^2\right ) \sqrt {1-k^2 x^2}}-\frac {1}{2 \sqrt {1-x^2} \left (1+k x^2\right ) \sqrt {1-k^2 x^2}}\right ) \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \left (1-k x^2\right ) \sqrt {1-k^2 x^2}} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (\sqrt {1-x^2} \sqrt {1-k^2 x^2}\right ) \int \frac {1}{\sqrt {1-x^2} \left (1+k x^2\right ) \sqrt {1-k^2 x^2}} \, dx}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} F\left (\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (-k;\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \Pi \left (k;\sin ^{-1}(x)|k^2\right )}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 10.26, size = 72, normalized size = 0.83 \begin {gather*} \frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (F\left (\text {ArcSin}(x)\left |k^2\right .\right )-\Pi \left (-k;\text {ArcSin}(x)\left |k^2\right .\right )-\Pi \left (k;\text {ArcSin}(x)\left |k^2\right .\right )\right )}{\sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + k^2*x^4)/(Sqrt[(1 - x^2)*(1 - k^2*x^2)]*(-1 + k^2*x^4)),x]

[Out]

(Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*(EllipticF[ArcSin[x], k^2] - EllipticPi[-k, ArcSin[x], k^2] - EllipticPi[k, A
rcSin[x], k^2]))/Sqrt[(-1 + x^2)*(-1 + k^2*x^2)]

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 1.63, size = 155, normalized size = 1.78

method	result	size
elliptic	\(\frac {\left (\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right )}{2+2 k}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right )}{-2+2 k}\right ) \sqrt {2}}{2}\)	\(87\)
default	\(\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticF \left (x , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , k , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}-\frac {\sqrt {-x^{2}+1}\, \sqrt {-k^{2} x^{2}+1}\, \EllipticPi \left (x , -k , k\right )}{\sqrt {k^{2} x^{4}-k^{2} x^{2}-x^{2}+1}}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((k^2*x^4+1)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x,method=_RETURNVERBOSE)

[Out]

(-x^2+1)^(1/2)*(-k^2*x^2+1)^(1/2)/(k^2*x^4-k^2*x^2-x^2+1)^(1/2)*EllipticF(x,k)-(-x^2+1)^(1/2)*(-k^2*x^2+1)^(1/
2)/(k^2*x^4-k^2*x^2-x^2+1)^(1/2)*EllipticPi(x,k,k)-(-x^2+1)^(1/2)*(-k^2*x^2+1)^(1/2)/(k^2*x^4-k^2*x^2-x^2+1)^(
1/2)*EllipticPi(x,-k,k)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((k^2*x^4+1)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x, algorithm="maxima")

[Out]

integrate((k^2*x^4 + 1)/((k^2*x^4 - 1)*sqrt((k^2*x^2 - 1)*(x^2 - 1))), x)

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Fricas [A]
time = 0.42, size = 80, normalized size = 0.92 \begin {gather*} \frac {{\left (k - 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k + 1\right )} x}\right ) + {\left (k + 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k - 1\right )} x}\right )}{2 \, {\left (k^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((k^2*x^4+1)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x, algorithm="fricas")

[Out]

1/2*((k - 1)*arctan(sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1)/((k + 1)*x)) + (k + 1)*arctan(sqrt(k^2*x^4 - (k^2 + 1)*x
^2 + 1)/((k - 1)*x)))/(k^2 - 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {k^{2} x^{4} + 1}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k x^{2} - 1\right ) \left (k x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((k**2*x**4+1)/((-x**2+1)*(-k**2*x**2+1))**(1/2)/(k**2*x**4-1),x)

[Out]

Integral((k**2*x**4 + 1)/(sqrt((x - 1)*(x + 1)*(k*x - 1)*(k*x + 1))*(k*x**2 - 1)*(k*x**2 + 1)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((k^2*x^4+1)/((-x^2+1)*(-k^2*x^2+1))^(1/2)/(k^2*x^4-1),x, algorithm="giac")

[Out]

integrate((k^2*x^4 + 1)/((k^2*x^4 - 1)*sqrt((k^2*x^2 - 1)*(x^2 - 1))), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {k^2\,x^4+1}{\left (k^2\,x^4-1\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((k^2*x^4 + 1)/((k^2*x^4 - 1)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)),x)

[Out]

int((k^2*x^4 + 1)/((k^2*x^4 - 1)*((x^2 - 1)*(k^2*x^2 - 1))^(1/2)), x)

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